or at least is bounded above and below by constant multiples of this and where $ \delta(x)$ is the distance from $ x $ to the boundary of $ \Omega$. Here $N$ is the space dimension of $\Omega$.

So using the integral representation and taking $ f(x) \ge 0 $ smooth and zero in a neighborhood of the boundary of $ \Omega$ i seem to be able to show that $ u(x)$ cannot be Holder continuous of order $> \frac{\alpha}{2}$ at the boundary. To do this let $ x_m$ be a sequence that converges to $x_0$ which lies on the boundary and assume that $ x_m$ approaches the boundary at right angles. Use the above representation to write out

$u(x_m)$ and note that one can calculate the maximum for big enough $m$ since $ f$ is identially zero near the boundary. Then one uses this to get a lower bound on the Holder quotient of $ u$ at $ x_m$ and $ x_0$ and arrives at a contradiction. (I will add more details of the exact calculation if this would help).

1 Answer
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I don't quite have the points to make this a comment like it should be (moderators, feel free to move), but there are multiple definitions of the fractional Laplacian, which might (as in, I don't know, not they definitely can) have different regularity properties at the boundary.

Which definition are you using? Can you give a reference for the Green's function or expound your calculations in more detail?